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kassandrarosario1115
12.02.2020 •
Engineering
A police officer in a patrol car parked in a 70 km/h speed zone observes a passing automobile traveling at a slow, constant speed. Believing that the driver of the automobile might be intoxicated, the office starts his car, accelerates uniformly to 90 km/h in 8 s, and, maintaining a constant velocity of 90 km/h, overtakes the motorist 42 s after the automobile passed him. Knowing that 18 s elapsed before the officer began pursuing the motorist, determine
(a) the distance the officer traveled before overtaking the motorist,
(b) the motorist's speed.
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Ответ:
S = 0.5 km
velocity of motorist = 42.857 km/h
Explanation:
given data
speed = 70 km/h
accelerates uniformly = 90 km/h
time = 8 s
overtakes motorist = 42 s
solution
we know initial velocity u1 of police = 0
final velocity u2 = 90 km/h = 25 mps
we apply here equation of motion
u2 = u1 + at
so acceleration a will be
a =![\frac{25-0}{8}](/tpl/images/0508/1782/d0dd3.png)
a = 3.125 m/s²
so
distance will be
S1 = 0.5 × a × t²
S1 = 100 m = 0.1 km
and
S2 = u2 × t
S2 = 25 × 16
S2 = 400 m = 0.4 km
so total distance travel by police
S = S1 + S2
S = 0.1 + 0.4
S = 0.5 km
and
when motorist travel with uniform velocity
than total time = 42 s
so velocity of motorist will be
velocity of motorist =![\frac{S}{t}](/tpl/images/0508/1782/7e104.png)
velocity of motorist =
velocity of motorist = 42.857 km/h
Ответ:
I don't understand...